Problem: Simplify and expand the following expression: $ \dfrac{3r - 3}{r + 6}+\dfrac{r}{3r + 2} $
Explanation: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(r + 6)(3r + 2)$ Multiply the first term by $\dfrac{3r + 2}{3r + 2}$ $ \begin{align*} \dfrac{3r - 3}{r + 6} \times \dfrac{3r + 2}{3r + 2} & = \dfrac{(3r - 3)(3r + 2)}{(r + 6)(3r + 2)} \\ & = \dfrac{9r^2 - 3r - 6}{(r + 6)(3r + 2)}\end{align*} $ Multiply the second term by $\dfrac{r + 6}{r + 6}$ $ \begin{align*} \dfrac{r}{3r + 2} \times \dfrac{r + 6}{r + 6} & = \dfrac{(r)(r + 6)}{(3r + 2)(r + 6)} \\ & = \dfrac{r^2 + 6r}{(3r + 2)(r + 6)}\end{align*} $ Now we have: $ = \dfrac{9r^2 - 3r - 6}{(r + 6)(3r + 2)} + \dfrac{r^2 + 6r}{(3r + 2)(r + 6)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{9r^2 - 3r - 6 + r^2 + 6r}{(r + 6)(3r + 2)} $ $ = \dfrac{10r^2 + 3r - 6}{(r + 6)(3r + 2)}$ Expand the denominator: $ = \dfrac{10r^2 + 3r - 6}{3r^2 + 20r + 12}$